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8/19/2019 PED 2 Report Group 8
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PROCESS EQUIPMENT
DESIGN –
II
ASSIGNMENT-I
SUBMITTED TO:
PROF. JAYANTA CHAKRABORTY
SUBMITTED BY:GROUP 8
RAHUL KUMAR 13CH10034
RAHUL YADAV 13CH10035
RAJA BARNWAL 13CH10036
RAKESH CHHIPA 13CH10037
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TABLE OF
CONTENTS
CONTENT PAGE
PROBLEM STATEMENT 03
BASIC THEORY 04
PROCESS DEISGN 05
MECHANICAL DESIGN 14
REFERENCES 21
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PROBLEM STATEMENT
A sieve plate column is to be designed to separate 8000 kg/hr of feedhaving 40 mole % Methanol (A) & 55 mole % Water (B) into an
overhead product containing 96 mole % A and a bottom product
containing mole 98 mole % B. The feed enters as an equilibrium
mixture of 30% liquid & 70% vapor. A reflux equal to 1.5 times the
minimum is to be used. Also an external reboiler is necessary to
remove the bottom product from the reboiler. The condensate
C & the
reflux enters the column at this temperature. Assume the Murphree
P ffiiy 70%. Gi αAB=3.91
(a) Complete the process design calculation.
(b) Mechanical design of the column.
(c) Enclose a drawing showing the details of the column.
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BASIC THEORY
Optimization stems from the need for improvement. Typical problems in
chemical engineering arise in process design, process control, model
development, process identification, and real-time optimization.
Distillation, being the most common industrial process for separation of
chemical components, is the focal unit of petroleum refining. While distillation
can be economically and easily scaled to different production levels, it is highly
energy intensive, consuming up to 80-90% of the total energy of a typical
petrochemical process. This makes the need for optimization essential to
maximize profitability of the entire process.
The performance of a distillation column is determined by many factors, such
as distillation feeds, internal liquid and fluid flow conditions, state of trays and
even ambient conditions which change over time, which makes the response
of real-time optimization to these changes a key contributor to successful
operation. In this particular process, detailed analysis of the operation and
design of the distillation column is performed involving the optimisation of
reflux ratio subject to various practical constraints and also the different
problems and cost analysis regarding the setting up of a distillation column inreal time.
In this design problem our task is to first find optimum value of reflux ratio at
which the total cost (operating + fixed cost) for setting up the distillation
facility for required separation of methanol is minimum. There are two
opposite factors contributing to the total cost. As the reflux ratio is increased
the number of trays required for a given degree of separation reduces which in
turn reduces the column height. However because of higher liquid and vapour
flow rates there is a increase in column diameter. Also the operating cost
increases because of higher heat duty of both the re-boiler and the condenser.
The optimum value of reflux ratio is determined by determining the minima in
a plot of Total cost Vs reflux ratio. The total cost calculated in our case is
actually annualised total cost which is calculated by dividing the total cost by
im “if” f qim f kig into account the interest
rates on various cost items.
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PROCESS DESIGN
In this problem, we need to separate methanol from a mixture of water and
methanol by using a bubble cap distillation column.
Given Data
Mol. Wt. (Methanol)(MWm) 32.04 gm/mol
Operating Pr. 14.7 Psig
Temperature (T) 148.5 F
Liquid Density (rhoL) 47.1645 lb/ft3
Vapour Density (rhoV) 0.08 lb/ft3
Liq. Surface Tension(sigma) 19.3 dyne/cm
Liquid Rate(L) 14875 lb/hr
Vapour Rate(V) 25581 lb/hr
Liquid rate (expressed in gpm) = 39.32 gpm
Vapour rate (expressed in cfs) = V*1/rhoV*3600
= (25581*1)/(.08*3600)
= 88.82292 cfs
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Cost optimization
F = 8000/(0.45*32+0.55*18) = 329.2181 kmol/hr
D = =(0.45-0.02)/(0.96-0.02)*F = 150.5998 kmol/hr
B = F-D = 178.6183 kmol/hr
Roptimum = Rmin * (R/Rmin)optimum
Refer to Excel sheet for detailed calculations. The optimization has been done
in MATLAB. The following is the result)
Roptimum = 1.3984
Theoretical number of trays have been derived from McCabe Thiele method.For calculation of column diameter, qc, qr, and cost of column, condenser,
reboiler, reboiler, cooling water and total annual cost, refer attached excel
sheet. Also refer [4]
Theoretical number of trays = 20
McCabe Thiele (Equilibrium curve and Equilibrium line)
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232000
233000
234000
235000
236000
237000
238000
239000
1 1.02 1.04 1.06 1.08 1.1 1.12
T o t a l A n n u a l C o s t
R/Rmin
Cost Optimization
R/Rmin |optimum = 1.0393
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Calculation of tower diameter:
F(lv) = (L /V)*sqrt(rhoV/rhoL)
= 0.023948418
Assume tray spacing = 24 inch
Cab= 0.4 [1](Figure 14.4)
Cab(corrected) = 0.4*(sigma/20)0.2
= 0.4*(19.3/20)0.2
= 0.397159956 (No measurable change)
Vapour velocity term (Un) = 0.4/(sqrt(rhoV/(rhoL-rhoV)))= 0.4/((sqrt(0.08/(47.1645-0.08)))
= 9.704071311 fps
Take flooding = 75%
Net tray area (An) = (1/0.75)*(88.82292/9.704071311)
= 12.20933562 ft2
Take 12% for downcomerColumn cross section area (At) = An/0.88
= 13.87424503 ft2
Dt = sqrt(4*At/pie)
= 4.203003381 ft
Next available standard diameter, Dn = 6 ft
From [1] Table 14.5, we get 24 inch tray spacing is suitable for a 6 ft diameter
tower
Area (new), An = pie* Dn2/4
= 28.27433388 ft2
From [1] Table 14.2, we get REVERSE FLOW
We take,
Riser area 0.1 * At
Down flow area 0.12 * At
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Max. active area 0.88 * At
Side weir length 0.62 * Dn
Max liq. flow path 0.3 * Dn
Dynamics slot submergence of 1.5 inch (Assumed)
Estimated tray pressure drop (ht)
= 0.53*(rhoV/rhoL)*(V/(0.1*An))+0.8*1.5+1
= 0.53*(0.08/47.1645)*(88.82292/(0.1*28.27433388))+0.8*1.5+1
= 2.228241208 inch of fluid
Tray Layout
Tower diameter, Dn = 6 ft
Tray spacing = 24 inch
Flow Type = Reverse Flow
Layout = Triangular
Bubble Cap Calculations
Bubble cap diameter = 4 inch
Area of 1 cap = pie*42/4
= 12.56637061 in2
Total bubble area, ABt = 0.7*An
= 19.79203372 ft2
Total number of bubble caps = ABt*144/area of 1 cap
= 226.8
Actual number of bubble caps = 227
Service is non-corrosive. We use CARBON STEEL – A 53B
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Summary (areas)
Area Formulae Per tray (ft2) % of An
Riser Area 4.8*227/144 7.566666667 26.76160895
Slot Area 8.12*227/144 12.80027778 45.27172181Downcomer Area 0.12*An 3.392920066 12
Active Area 0.88*An 24.88141382 88
Area under apron Area under baffle 0.5 1.768388257
Tower area An 28.27433388 100
Net area 0.88*An 24.88141382 88
Flooding
Vapour velocity based on net area, U = V/An
= 88.82292 cfs/24.88141382 ft2
= 3.569850062 ft/s
% flooding = U/Un = 36.787 % (Under limits)
Entrainment
Phi = 0.055 [1] (Figure 14.5)
It is well within limits
Shape factor = Rs = 0.167/0.333 = 0.501501502
For Rs = 0.50, we have Cs = 0.74 [1] (Figure 14.6)
Q max = Cs*Slot area*sqrt(slot height*(rhoL-rhoV)/rhoV)
= 229.7973954 cfs
Vapour load = 100*V/Q max = 38.65 % of slot capacity
Slot opening = 55% [1] (Figure)
Slot height = 1*0.55 = 0.55 inch
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Liquid crest over weir
Liquid load = 39.32 gpm
HOW = 0.48*(L/lw)2/3
= 0.51 inch
From Figure 14.7 [1],
L/lw2.5
= 2.2522
Weir correction = Fs = 1.04
HOW (corrected) = Fs*HOW = 0.53 inch
Liquid gradient
Arithmetic average flow width = 2.5 ft (approximated)Liquid load per unit width = L/2.5 = 15.728 gpm/ft
CD = 0.75 [1] (Figure 14.9 and 14.10)
Hl = 2.75+0.53+1 = 5 (approximated)
Del = 0.5 inch (using figure 14.9 and 14.10)
Vapour pressure Drop
hcd = 0.53*(0.08/47.1645)*(D15/(0.1*B30)) = 0.03 inch of liquid
Mean dynamic slot seal, hda = 0.5+0.53+0.5/2 = 1.28 inch of liquid
Fva = 1.62 [1](Equation 14.40)
hal = 0.6*hda = 0.98 inch of liquid [1](figure 14.15 and equation 14.37)
Total tray pressure drop, ht = 0.03+0.51+0.98 = 1.52 inch of liquid
Vapour distribution Ratio
hc = 0.03+0.51 = 0.54 inch of liquid
Rcd = 0.5/0.54 = 0.925
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Downcomer dynamics
had = 0.2 inch of liquid [1](using Equation 14.41 and 14.42)
hdc = 6.2 inch of liquid [1](using Equation 14.43)
hfd = 6.2/.4 =15.5 inch of liquid
tdc = 3.8 seconds [1](using Equation 14.47)
TRAY AND BUBBLE CAP DESIGN (all dimensions in mm)
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MECHANICAL DESIGN
Shell ID (Di) 1828.8 mm
No. of trays 20
Tray spacing (t) 609.6 mm
Hole Diameter (dh) 101.6 mm
Plate Thicknes (tp) 101.6 mm
Weir height (hw) 69.85 mm
Material for tray Carbon Steel A 53B
A 53BShell material Carbon Steel
Allowable stress for shell mat.(Pa) 1.03E+08 PaDensity of shell mat 7850 kg/m3
skirt height 25.4 Mm
Operating pr (p) 112405 Pa
Design pr (pD) 123645.5 Pa
Operating temp 64.7 to 100 *C
Design temp 110 *C
Top disengaging space 609.6 mm
Bottom separator space 1000 mmInsulation material taken Asbestos
Insulation thickness considered 50 mm
Density of insulation 270 kg/m3
Formulae source: [2] and [6]
Shell thickness
J = 0.8
Minimum shell thickness, ts = p*Dt/(2*f*J-p)
= 1.367819 mm
After adding corrosion allowance, ts = 8mm
Do = Di+2*ts
= 1844.8 mm
th = pD*Di/(2*J*Pa-0.2*pD)
= 1.367001 mm
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Adding corrosion allowance, th = 8 mm
Column height = (t + 19*t + 20*tp + 1000)/1000 = 15.224 m
(Refer to Excel sheet along with this file for the calculations)
Stresses developed in column
Formulae
Source [2]
Axial Stress (f as) 8910204 Pa Eq. 9.3.1
Circumferential Stress (f cs) 17820408 Pa Eq. 9.3.1
Compressive stresses caused by dead loads
Wshell 54066.25 N
Mshell 6620.357 kg
f(dead wt shell) 29434.89 Pa Eq. 9.3.2
M(insulation) 1191.137 kg
f(dead wt insulation) 955285 Pa Eq. 9.3.3
Active Area 2.31 m2
Liquid depth on trays 83.2977 mm
Mass of liquid 5002.86 Kg
f(dead wt liquid) 4012265 Pa Eq. 9.3.4
Stress induced by column attachments
f(dead wt attachments) 5309488 Pa Eq. 9.3.5
f(dead total) 10306473 Pa Eq. 9.3.6
f(dead total)
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Below the guy ring
(Formulae used, [2] Eq. 9.3.8-9.4.4)
M(w,H,guy)(max) 26723.1 N
f(wind,H,below guy) 5269799 Pa
f(com guy) 641969.5 Pa
Analysis of stress
For upwind side
Total stress 82920516 Pa
For downwind side
Total stress 92472689 Pa
Torispherical Head has a range of 0.1 to 1.5 MPa. It is, therefore,
suitable to use for our column design.
Design of head
Do 1844.8 mm
ro 110.688 mm 0.06*Do
ho 312.394 mm Eq. 4.2.22
Do2/4*Ro 461.2 mm
(Do*ro/2)^0.5 319.5287 mm
hE, the minimum of above 3 312.394 mm
hE/Do 0.169338 mm
As the diameter of the vessel is not very large, head can be fabricated
from single plate, therefore J = 1
t/DoC 0.000544
C 3.679807 Table 4.1A
t/Do 0.002
t 3.689856 mm Eq. 4.2.20Corrosion allowance 2 mm
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Uncorroded plate thickness 6 mm
Torispherical Head Volume 0.623133 m3
Skirt Support Design
Mshell 6620.357 Kg
Wshell 64879.5 N
Desity of Carbon Steel 7850 Kg/m3
WD 47937.64 N
Wmin 112817.1 NWi 11673.14 N
Wl 391901.7 N
Wmax 516392 N
Period of vibration at min dead wt
Tmin 0.178766 s
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Tmax 0.382461 s
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Design of skirt bearing plate
Stress_max(compressive) 0.099138 MN/m2
Thickness of bearing plate, tbp 5.363147 mm
According to IS code [3] 7 mm
As thickness is
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Area available from shell reinforcrment
As 0.003650572 m2
Area available from nozzle
reinforcrment, An 2H1(tn-tr'-c)
tr' 0.004160723 m
H1 0.05396295 m
An 0.000953987 m2
Reinforcement area available from shell and
nozzle is
As+An 0.004604559 m2
Area remained to be
compansated
A-(As+An) 0.00227287 m2
Ar >= 0.00227287 m2
tp 0.010538229
Ring Pad dimension
Inner Diameter, do 0.25 m
Outer diameter 0.448 m
Thickness 0.025 m
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REFERENCES
[1] Smi, B.D., “Dig f Eqiibim Sg P”, MGw-Hill Book Co.
[2] Introduction to Chemical Equipment Design by B.C Bhattacharya, 2012
[3] IS : 2825-1969, the Indian Standard codes
[4] Plant Design and Economics for Chemical Engineers Fourth Edition
by Max S. Peters & Klaus D. Timmerhaus
[5] R.H.Py, D.W.G; Py’ Cmi Egi’ Hbk7 Eii
[6] R. K. Si, C Ri’ Chemical Engineering: Chemical
Engineering Design (vol. 6), Butterworth-Heinemann, 3rd ed. 1999